|
%I #44 Feb 22 2024 10:40:58
%S 1,3,10,41,190,946,4940,26693,147990,837102,4811860,28027210,
%T 165057100,981177060,5879570200,35478788269,215398416870,
%U 1314794380374,8064119033220,49673222082782,307163049317540,1906066361809148,11865666767361960,74081851132379426
%N Certain directed lattice paths.
%C See Dziemianczuk (2014) for precise definition.
%H Lars Blomberg, <a href="/A260772/b260772.txt">Table of n, a(n) for n = 0..100</a>
%H M. Dziemianczuk, <a href="http://arxiv.org/abs/1410.5747">On Directed Lattice Paths With Additional Vertical Steps</a>, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
%H M. Dziemianczuk, <a href="https://doi.org/10.1016/j.disc.2015.11.001">On Directed Lattice Paths With Additional Vertical Steps</a>, Discrete Mathematics, Volume 339, Issue 3, 6 March 2016, Pages 1116-1139.
%H Heba Bou KaedBey, Mark van Hoeij, and Man Cheung Tsui, <a href="https://arxiv.org/abs/2402.11121">Solving Third Order Linear Difference Equations in Terms of Second Order Equations</a>, arXiv:2402.11121 [math.AC], 2024. See p. 3.
%F G.f.: P1(x) = (2*(1-x)/3)/x - ((2*sqrt(1-5*x-2*x^2)/3)/x)*sin((Pi/6 + arccos(((20*x^3-6*x^2+15*x-2)/2)/(1-5*x-2*x^2)^(3/2))/3)). - See Dziemianczuk (2014), Proposition 11.
%F a(n) = (1/n)*Sum_{j=0..(n+1)/4} (-1)^j*C(n,j)*C(3*n-4*j,n-4*j+1), a(0)=1. - _Vladimir Kruchinin_, Apr 04 2019
%F n*(n+1)*(25*n^2-70*n+21)*a(n) - 30*(7*n-15)*n*a(n-1) + (-1100*n^4+5280*n^3-6424*n^2-1188*n+3816)*a(n-2) + 120*(n+2)*(n-3)*a(n-3) - 16*(n-3)*(n-4)*(25*n^2-20*n-24)*a(n-4) = 0. - _Mark van Hoeij_, Jul 14 2022
%F a(n) ~ 2^(n - 1/2) * phi^((10*n - 1)/4) / (sqrt(Pi) * 5^(1/4) * sqrt(phi^(3/2) - 2) * n^(3/2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Jul 15 2022
%p # A260772 satisfies a 4th-order recurrence that can be reduced
%p # to a 2nd-order recurrence given in this program t:
%p t := proc(n) options remember;
%p if n <= 1 then
%p [-1/2, 0, 1, 4][2*n+2]
%p else
%p (16*(n-2)*(2*n-3)*(5*n-2)*t(n-2) + (440*n^3-1056*n^2+724*n-144)*t(n-1))
%p /( n*(2*n+1)*(5*n-7) )
%p fi
%p end:
%p A260772 := proc(n)
%p t(n/2) + ( (2-2*n)*t((n-1)/2)+(n+2)*t((n+1)/2) ) / (1+5*n)
%p end:
%p seq(A260772(i),i=0..100);
%p # _Mark van Hoeij_, Jul 14 2022
%o (Maxima)
%o a(n):=if n=0 then 1 else sum((-1)^j*binomial(n,j)*binomial(3*n-4*j,n-4*j+1),j,0,(n+1)/4)/n; /* _Vladimir Kruchinin_, Apr 04 2019 */
%o (PARI) a(n) = if (n==0, 1, sum(j=0, (n+1)/4, (-1)^j*binomial(n,j)*binomial(3*n-4*j, n-4*j+1))/n); \\ _Michel Marcus_, Apr 05 2019
%Y Cf. A122868, A260774, A064641, A156016.
%Y Cf. A006139, A179191, A052709, A025227.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Jul 30 2015
%E More terms from _Lars Blomberg_, Aug 01 2015
| 